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The classical Caratheodory theorem states that if $G$ is a simply connected domain in the plane, whose boundary is a Jordan curve, then the Riemann uniformization extends continuously to a homeomorphism between the boundary of $G$ and the unit circle.

Question: does this hold if the domain is embedded in a compact Riemann surface of higher genus? More precisely, is it true that if $G$ is a simply connected domain in a compact Riemann surface $X$, and if the boundary of $G$ is a simple closed curve, then the uniformizing map extends as a homeomorphism between the boundary of $G$ and the unit circle?

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    $\begingroup$ The uniformization theorem applied to $X$ gives that its universal cover is one of $\Bbb P^1, \Bbb C, $ or the unit disc. The compact simply connected domain $G$ lifts to the universal cover, which exhibits it as a subset of $\Bbb C$. So this is just the classical Caratheodary theorem. $\endgroup$
    – mme
    Commented Jan 20, 2020 at 14:30
  • $\begingroup$ @MikeMiller I thought of that, but a priori lifting to the universal cover will change the boundary. Is there a simple argument why the closure of $G$ would lift to the closure of a Jordan domain? $\endgroup$
    – Albert
    Commented Jan 20, 2020 at 14:49
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    $\begingroup$ I said "compact domain" for a reason --- you should include the boundary of $G$ when talking about lifting it, so that $G$ is homeomorphic to a closed unit disc. Then the whole closed unit disc lifts to the universal cover, giving a closed unit disc embedded upstairs (the boundary lifting the boundary, the interior lifting the interior). $\endgroup$
    – mme
    Commented Jan 20, 2020 at 15:01
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    $\begingroup$ @MikeMiller I see, but if I may ask one more stupid question: how do you know that the closure of $G$ is homeomorphic to the closed unit disk? this seems sort of close to proving what we want (keeping in mind that in general the closure of a simply connected domain needs not be simply connected, of course) $\endgroup$
    – Albert
    Commented Jan 20, 2020 at 15:14
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    $\begingroup$ This seems trickier than I thought at first, but still not too bad. A proof uses that any embedded curve $C \subset S$ in a surface has, for any point $p \in C$, a neighborhood $U$ and a homeomorphism $f: U \to \Bbb R^2$ with $f(C \cap U) = \Bbb R$ and $f(p) = 0$. Then $f(G \cap U)$ is an open set in $\Bbb R^2$ with boundary $\Bbb R$. Some fiddling shows that it must either be one open half-plane or the union of both; a little more shows that whether it's one half-plane or two is independent of the point $p \in C$. But if it was both, then we would have $G = S \setminus C$, which is not s.c. $\endgroup$
    – mme
    Commented Jan 20, 2020 at 16:01

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